Q.The rate law equation for A → Product, is rate = k[A]$^x$. What is the effect of increase in concentration of 'A' on rate of reaction, if x < 0?

A rate constant with units of s$^{-1}$ (per second) indicates a first-order reaction.

The unit of the rate constant $k$ for a reaction of order $n$ is $mol^{1-n}\,L^{n-1}\,s^{-1}$ (concentration$^{1-n}$ time$^{-1}$).

• Zero order: $mol\,L^{-1}\,s^{-1}$
• First order: $s^{-1}$ (concentration terms cancel out) …

With $x<0$, Rate $=k[A]^x$ falls as $[A]$ increases — the rate decreases.

Given rate law: $\text{Rate} = k[A]^x$.

If the order $x$ is negative (say $x=-1$), then $\text{Rate} = \dfrac{k}{[A]}$ — the rate is inversely related to $[A]$.

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Rate-constant unit $mol\,dm^{-3}s^{-1}$ corresponds to order $n=0$.

The general rate-constant unit is $\text{mol}^{1-n}\,dm^{3(n-1)}\,s^{-1}$ for a reaction of order $n$.

Given unit: $mol\,dm^{-3}\,s^{-1} = mol^{1}\,dm^{-3}\,s^{-1}$, matching $n=0$ (since $1-n=1 \Rightarrow n=0$).

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$2 = 4^n \Rightarrow n = 0.5$.

For $A \rightarrow$ products, rate $= k[A]^n$. If concentration is increased by a factor of 4 and the rate only doubles:

$\dfrac{\text{rate}_2}{\text{rate}_1} = \left(\dfrac{[A]_2}{[A]_1}\right)^n \Rightarrow 2 = 4^n$ …

$k = \text{rate}/[A] = 1.1\times10^{-4}\ s^{-1}$.

For a first order reaction $A \rightarrow B$, rate $= k[A]$.

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